(1)
Photo-electrons with

*minimum velocity*imply that**K**applies here_{min }**so indicates the wavelength of incident light (l = 0.50 mm) is***so: l = l*__at the cutoff wavelength___{c }Then the*cutoff or*is:_{ }threshold frequency
f

_{ c}= c /l_{c}= (3 x 10^{8}ms^{-1})/ (500 x 10^{-9}m) = 6 x 10^{14 }s^{-1}
Then
work function would be: f = hf

(6.62 x 10

_{ c}=(6.62 x 10

^{-34}Js) ( 6 x 10^{14 }s^{-1})^{ }
= 3.97 x 10

^{-19}J = (3.97 x 10^{-19}J )/ 1.60 x 10^{-19}J /eV) = 2.48 eV
(2)
Stopping potential V

**= 0.54 V and l = 750 nm**_{s}
f= c/ l =
(3 x 10

^{8}ms^{-1})/ (750 x 10^{-9}m) = 4 x 10^{14 }s^{-1}
The
work function is obtained from: eV

**hf - f**_{s}=
So,
transposing: f = hf
- eV

**=**_{s }
(6.62 x 10

^{-34}Js) (4 x 10^{14 }/s) - (1.6 x 10^{-19}C) 0.54 J/C]
f = 2.64 x 10

^{-19}J - 0.86 x 10^{-19}J = 1.78 x 10^{-19}J
In
eV: f
= 1.78 x 10

^{-19}J / (1.6 x 10^{-19}J/ eV) = 1.11 eV
(3)
K

_{max }= 0.57 eV and photo-electrons dislodged from a metal surface by incident radiation with l = 500 nm.
The
incident energy E = hf = h c/l =

(6.62 x 10

(6.62 x 10

^{-34}Js) (3 x 10^{8}ms^{-1})/ (500 x 10^{-9}m)
h c/ l = 3.97 x 10

^{-19}J**K**0.57 eV (1.6 x 10

_{max }=^{-19}J/ eV) = 0.91 x 10

^{-19}J

Therefore:

**K**0.91 x 10

_{max }=^{-19}J

= [3.97 x 10

^{-19}J - f ]
So,
the work function is: f =
[3.97 x 10

^{-19}J - 0.91 x 10^{-19}J] = 3.06 x 10^{-19}J
To
get in eV: (3.06 x 10

^{-19}J)/ (1.6 x 10^{-19}J/ eV) = 1.91 eV
The
stopping potential in volts:

V

**hf/ e - f/e where the slope h/e = 4.13 x 10**_{s}=^{-15}Js/C
The
frequency f = c/l = (3
x 10

6 x 10

^{8}ms^{-1})/ (500 x 10^{-9}m) =6 x 10

^{14 }c/s
Then:

V

(3.06 x 10

**(4.13 x 10**_{s}=^{-15}J s/C )(6 x 10^{14 }s^{-1}) -(3.06 x 10

^{-19}J)/ (1.6 x 10^{-19}C)
V

**= 2.47 V – 1.91 V = 0.56 V**_{s}
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